What is the first derivative and second derivative of f(x) = (x-1)^2 (x-3)?

Jun 27, 2018

$f ' \left(x\right) = 3 {\left(x - 1\right)}^{2} - 4 \left(x - 1\right)$

$f ' ' \left(x\right) = 6 x - 10$

Explanation:

Let $u = x - 1 , x - 3 = u - 2 \therefore \frac{\mathrm{du}}{\mathrm{dx}} = 1$

${u}^{2} \left(u - 2\right) = {u}^{3} - 2 {u}^{2}$

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}}} \left({u}^{3} - 2 {u}^{2}\right)$

$3 {u}^{2} \cdot \frac{\mathrm{du}}{\mathrm{dx}} - 4 u \frac{\mathrm{du}}{\mathrm{dx}}$

Substitute

$f ' \left(x\right) = 3 {\left(x - 1\right)}^{2} - 4 \left(x - 1\right)$

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}}} \left(3 {u}^{2} - 4 u\right)$

$6 u \frac{\mathrm{du}}{\mathrm{dx}} - 4 \frac{\mathrm{du}}{\mathrm{dx}}$

Substitute

$f ' ' \left(x\right) = 6 \left(x - 1\right) - 4 = 6 x - 10$